jmc

We calculate the probability that he does not draw a card from at least three of the suits. To do this, we calculate the number of sets of 5 cards from at most two suits and divide by $\left(\genfrac{}{}{0ex}{}{52}{5}\right)$, the number of sets of 5 cards. Since there are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ choices for the two suits, and $\left(\genfrac{}{}{0ex}{}{26}{5}\right)$ ways to choose 5 cards from the 26 in those two suits, our answer would appear to be $6\left(\genfrac{}{}{0ex}{}{26}{5}\right)$. But this triple-counts the ways to choose the cards from a single suit: 5 hearts is included in 5 (hearts and spades), 5 (hearts and clubs), and 5 (hearts and diamonds). So we subtract twice the number of ways to choose cards from a single suit: $6\left(\genfrac{}{}{0ex}{}{26}{5}\right)-2\cdot 4\left(\genfrac{}{}{0ex}{}{13}{5}\right)$. We divide this by $\left(\genfrac{}{}{0ex}{}{52}{5}\right)$ to get $$\frac{6\cdot 26\cdot 25\cdot 24\cdot 23\cdot 22-8\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9}{52\cdot 51\cdot 50\cdot 49\cdot 48}=\frac{88}{595}.$$Therefore, the probability that he draws three or four of the suits is $1-\frac{88}{595}=\boxed{\frac{507}{595}}$.